APPENDIX C
C.2
Differential Equations
A27
Separation of Variables
Use separation of variables to solve differential equations. • Use differential equations to model and solve real-life problems.
Separation of Variables
The simplest type of differential equation is one of the form y f x. You know
that this type of equation can be solved by integration to obtain
y
f x dx.
In this section, you will learn how to use integration to solve another important
family of differential equations—those in which the variables can be separated.
This technique is called separation of variables.
Separation of Variables
If f and g are continuous functions, then the differential equation
dy
f xg y
dx
1
dy g y
f x dx C.
Essentially, the technique of separation of variables is just what its name
implies. For a differential equation involving x and y, you separate the x variables
to one side and the y variables to the other. After separating variables, integrate
each side to obtain the general solution. Here is an example.
EXAMPLE 1
Solving a Differential Equation
Find the general solution of
dy
x
.
dx y2 1
Solution Begin by separating variables, then integrate each side.
dy
x
2
dx y 1
Differential equation
(y 2 1 dy x dx
Separate variables.
y 2 1 dy Integrate each side.
You can use a symbolic
integration utility to solve
a separable variables differential
equation. Use a symbolic integration utility to solve the differential
equation
y has a general solution of
TECHNOLOGY
x dx
y3
x2
y C
3
2
General solution
x
.
y2 1
A28
APPENDIX C
Differential Equations
EXAMPLE 2
Solving a Differential Equation
Find the general solution of
dy x
.
dx y
Solution Begin by separating variables, then integrate each side.
dy x
dx y
Differential equation
y dy x dx
Separate variables.
y dy Integrate each side.
x dx
y2 x2
C1
2
2
Find antiderivatives.
y2 x2 C
Multiply each side by 2.
So, the general solution is y 2 x2 C. Note that C1 is used as a temporary
constant of integration in anticipation of multiplying each side of the equation by
2 to produce the constant C.
STUDY TIP
After finding the general solution of a differential equation, you should use the
techniques demonstrated in Section C.1 to check the solution. For instance, in
Example 2 you can check the solution by differentiating the equation
y 2 x 2 C to obtain 2yy 2x or y xy.
EXAMPLE 3
Solving a Differential Equation
Find the general solution of
ey
dy
2x.
dx
Use a graphing utility to graph several solutions.
Solution Begin by separating variables, then integrate each side.
ey
5
C
C
5
15
C
6
10
6
dy
2x
dx
Differential equation
e y dy 2x dx
Separate variables.
e y dy Integrate each side.
2x dx
e y x2 C
Find antiderivatives.
By taking the natural logarithm of each side, you can write the general solution as
C
0
y lnx2 C.
5
FIGURE A.10
General solution
The graphs of the particular solutions given by C 0, C 5, C 10, and
C 15 are shown in Figure A.10.
APPENDIX C
EXAMPLE 4
Differential Equations
A29
Finding a Particular Solution
Solve the differential equation xe x yy 0 subject to the initial condition
y 1 when x 0.
2
Solution
xe x yy 0
2
y
Differential equation
dy
2
xe x
dx
2
Subtract xe x from each side.
y dy xe x dx
Separate variables.
y dy Integrate each side.
2
xe x dx
2
y2
1 2
e x C1
2
2
Find antiderivatives.
y 2 e x C
Multiply each side by 2.
2
To find the particular solution, substitute the initial condition values to obtain
12 e0 C.
2
This implies that 1 1 C, or C 2. So, the particular solution that satisfies
the initial condition is
y2 e x 2.
2
EXAMPLE 5
Particular solution
Solving a Differential Equation
Example 3 in Section C.1 uses the differential equation
dx
k10 x
dt
to model the sales of a new product. Solve this differential equation.
Solution
dx
k10 x
dt
STUDY TIP
Differential equation
1
dx k dt
10 x
Separate variables.
1
dx 10 x
Integrate each side.
k dt
ln10 x kt C1
ln10 x kt C1
10 x ektC1
x 10 Cekt
Find antiderivatives.
Multiply each side by 1.
Exponentiate each side.
Solve for x.
In Example 5, the context of
the original model indicates that
10 x is positive. So, when
you integrate 110 x, you
can write ln10 x, rather
than ln 10 x .
Also note in Example 5
that the solution agrees with
the one that was given in
Example 3 in Section C.1.
A30
APPENDIX C
Differential Equations
Applications
EXAMPLE 6
Modeling National Income
Let y represent the national income, let a represent the income spent on necessities, and let b represent the percent of the remaining income spent on luxuries. A
commonly used economic model that relates these three quantities is
dy
k1 b y a
dt
where t is the time in years. Assume that b is 75%, and solve the resulting differential equation.
Corporate profits in the United
States are closely monitored by
New York City’s Wall Street executives. Corporate profits, however,
represent only about 11.9% of
the national income. In 1999, the
national income was more than
$7 trillion. Of this, about 71%
was employee compensation.
Solution Because b is 75%, it follows that 1 b is 0.25. So, you can solve the
differential equation as follows.
dy
k0.25 y a
dt
Differential equation
1
dy 0.25k dt
ya
Separate variables.
1
dy ya
Integrate each side.
0.25k dt
ln y a 0.25kt C1
y a Ce0.25kt
y a Ce0.25kt
Find antiderivatives, given y a > 0.
Exponentiate each side.
Add a to each side.
The graph of this solution is shown in Figure A.11. In the figure, note that the
national income is spent in three ways.
National income necessities luxuries capital investment
Modeling National Income
y
Capital investment
Consumed on luxuries
Consumed on necessities
Income consumed on
necessities and luxuries
y = a + 0.75Ce 0.25kt
National income
y = a + Ce 0.25kt
C
a
t
1
2
FIGURE A.11
3
4
5
6 7 8 9
Time (in years)
10 11 12 13 14
APPENDIX C
EXAMPLE 7
A31
Differential Equations
Using Graphical Information
Find the equation of the graph that has the characteristics listed below.
1. At each point x, y on the graph, the slope is x2y.
2. The graph passes through the point 2, 1.
Solution Using the information about the slope of the graph, you can write the
differential equation
x
dy
.
dx
2y
Using the point on the graph, you can determine the initial condition y 1 when
x 2.
dy
x
dx
2y
Differential equation
2y dy x dx
Separate variables.
2y dy x dx
Integrate each side.
x2
C1
2
Find antiderivatives.
y2 2y 2 x 2 C
x 2y 2 C
2
Multiply each side by 2.
Simplify.
2
Applying the initial condition yields
22 212 C
3
3
which implies that C 6. So, the equation that satisfies the two given conditions
is
x 2 2y 2 6.
As shown in Figure A.12, the graph of this equation is an ellipse.
TAKE ANOTHER LOOK
Classifying Differential Equations
In which of the differential equations can the variables be separated?
a.
dy 3x
dx
y
b.
dy 3x
1
dx
y
c. x 2
d.
dy 3x
dx
y
dy 3x y
dx
y
2
Particular solution
FIGURE A.12
A32
APPENDIX C
Differential Equations
The following warm-up exercises involve skills that were covered in earlier sections.
You will use these skills in the exercise set for this section.
WA R M - U P C . 2
In Exercises 1–6, find the indefinite integral and check your result by differentiating.
1.
2.
3.
4.
5.
6.
x 32 dx
t3 t13 dt
2
dx
x5
y
dy
2y 2 1
e2y dy
2
xe1x dx
In Exercises 7–10, solve the equation for C or k.
7. 32 63 1 C
8. 12 22 C
10. 62 36 ek
9. 10 2e2k
EXERCISES C.2
In Exercises 1–6, decide whether the variables in the differential
equation can be separated.
dy
x
1.
dx y 3
dy x 1
2.
dx
x
3.
dy 1
1
dx
x
4.
x
dy
dx x y
5.
dy
xy
dx
6. x
dy 1
dx
y
In Exercises 7–26, use separation of variables to find the general
solution of the differential equation.
7.
dy
2x
dx
9. 3y 2
8.
dy
1
dx
11. y 1
dy
2x
dx
13. y xy 0
et
10.
dy 1
dx
x
dy
x 2y
dx
12. 1 y
dy
4x 0
dx
14. y y 5
15.
dy
dt
4y
16. e y
17.
dy
1 y
dx
18.
dy
3t2 1
dt
dy
dx
xy
19. 2 xy 2y
20. y 2x 1 y 3
21. xy y
22. y yx 1 0
23. y x
x
y 1y
25. e x y 1 1
24.
dy x 2 2
dx
3y 2
26. yy 2xe x 0
In Exercises 27–32, use the initial condition to find the particular
solution of the differential equation.
Differential Equation
Initial Condition
27. yy e 0
y 4 when x 0
28. x y y 0
y 4 when x 1
29. x y 4 y 0
y 5 when x 0
x
30.
dy
x 21 y
dx
y 3 when x 0
31. dP 6P dt 0
P 5 when t 0
32. dT kT 70 dt 0
T 140 when t 0
APPENDIX C
In Exercises 33 and 34, find an equation for the graph that passes
through the point and has the specified slope. Then graph the
equation.
33. Point: 1, 1
Slope: y 6x
5y
41. Learning Theory The management of a factory has
found that a worker can produce at most 30 units per day.
The number of units N per day produced by a new employee will increase at a rate proportional to the difference
between 30 and N. This is described by the differential
equation
dN
k30 N
dt
34. Point: 8, 2
Slope: y 2y
3x
Velocity In Exercises 35 and 36, solve the differential equation
to find velocity v as a function of time t if v 0 when t 0. The
differential equation models the motion of two people on a
toboggan after consideration of the force of gravity, friction, and
air resistance.
35. 12.5
dv
43.2 1.25v
dt
36. 12.5
dv
43.2 1.75v
dt
Chemistry: Newton’s Law of Cooling In Exercises 37–39,
use Newton’s Law of Cooling, which states that the rate of change
in the temperature T of an object is proportional to the difference
between the temperature T of the object and the temperature T0
of the surrounding environment. This is described by the differential equation dTdt kT T0.
37. A steel ingot whose temperature is 1500F is placed in a
room whose temperature is a constant 90F. One hour later,
the temperature of the ingot is 1120F. What is the ingot’s
temperature 5 hours after it is placed in the room?
38. A room is kept at a constant temperature of 70F. An object
placed in the room cools from 350F to 150F in 45 minutes. How long will it take for the object to cool to a
temperature of 80F?
39. Food at a temperature of 70F is placed in a freezer that is
set at 0F. After 1 hour, the temperature of the food is 48F.
where t is the time in days. Solve this differential equation.
42. Sales The rate of increase in sales S (in thousands of
units) of a product is proportional to the current level of
sales and inversely proportional to the square of the time t.
This is described by the differential equation
dS kS
2
dt
t
where t is the time in years. The saturation point for the
market is 50,000 units. That is, the limit of S as t → is
50. After 1 year, 10,000 units have been sold. Find S as a
function of the time t.
43. Economics: Pareto’s Law According to the economist Vilfredo Pareto (1848–1923), the rate of decrease of
the number of people y in a stable economy having an
income of at least x dollars is directly proportional to the
number of such people and inversely proportional to their
income x. This is modeled by the differential equation
dy
y
k .
dx
x
Solve this differential equation.
44. Economics: Pareto’s Law In 1998, 8.6 million people
in the United States earned more than $75,000 and 50.3
million people earned more than $25,000 (see figure).
Assume that Pareto’s Law holds and use the result of
Exercise 43 to determine the number of people (in millions) who earned (a) more than $20,000 and (b) more than
$100,000. (Source: U.S. Census Bureau)
(a) Find the temperature of the food after it has been in the
freezer 6 hours.
40. Biology: Cell Growth The rate of growth of a spherical cell with volume V is proportional to its surface area S.
For a sphere, the surface area and volume are related by
S kV 23. So, a model for the cell’s growth is
dV
kV 23.
dt
Solve this differential equation.
Pareto’s Law
y
Number of people (in millions)
(b) How long will it take the food to cool to a temperature
of 10F?
A33
Differential Equations
150
100
50
(25,000, 50.3)
(75,000, 8.6)
x
50,000
100,000
150,000
Earnings (in dollars)
200,000